The interior of a connected set in $\mathbb R^k$

No. If $X\subset\mathbb R^2$ is the union of two closed disks of radius $1$, one with center at $(1;0)$ and another with center at $(-1;0)$, then $X$ is connected but its interior is not.

Nope... Pick two tangent balls...

No. Let

$A_1=\{(x,y)|x\leq0,y\leq0\}$ (The third quadrant and the positive x-axis) and

$A_2=\{(x,y)|x\geq0,y\geq0\}$.(The first quadrant and the positive x-axis).

$A_1\cup A_2$ is connected. Their interior is the first and third quadrants.

Proof by visualization.